Im going to assume you've never played Roulette in a Casino before have you?Placing multiple bets is the only way to seemingly beat the house.With nothing but singular bets, the house will theoretically always win.

Say i bet on the 1st collumThats 33% win 64%loose with 33%=300%If i bet on 1st collum and 1st-12Then it becomes around 44% to win 300% 56%loose With a chance to win 600% of my bet because i can win money from both bets from One spin

Roulette is based upon the ability to bet multiple times. Because say i loose on that double bet, I lost Twice my regular bet. But if i hit that 600% then i make back all the money and then double the bet (200% to pay for the bet 400%profit)\

I've played roulette on online casinos plenty of times, but never in a real casino.

Your logic makes no sense, as the house edge remains the same regardless of how many bets you place. I understand how you're thinking, but in the end, return remains the same. Your chance of getting that 600% win is no better than getting the 300% win two times. In the long run, the odds remain the same. The only thing that changes with multiple bets is variance, return and house edge ultimately remains the same.

This is why roulette is such a profitable game. You can speculate on the odds, but you never really know what they are...except not in your favor.

You know exactly what the odds are.

The bet you can make with the highest chance of winning is on a black or red column. There are 36 numbers along with 0 and sometimes 00. So you have a 48.64% chance of hitting the colored column you bet on with a single 0 wheel, but the payout is only double your bet. The extra 2.72% is the house's cut, on average.

Likewise, if you bet on a single number, you have a 2.70% chance of hitting that number. But the payout is only 36 times your bet amount, or 97.29% on average. The extra 2.72% is the house's cut, on average.

If the house uses a double green wheel, then you double the house's cut.

Indeed--I guess my comment was meant in a broader perspective. The game itself is designed to manipulate these percentages, like the 2.72 cut from the house or a double green. You can know the odds and play the odds and still lose.

I still don't get what you're trying to say... of course you'll lose - the odds are always against you!

Yes, we talked about Roulette in the German Area before. Right now you can bring some coins to the Table and Bet on As many Fields As you like at the Same Time. You have different Chips starting at 0.1 bitcoin. The Maximum Single Chip at the Moment is 1 btc. But that would Be only Limited by the Bank Balance. I'll keep you informed, and maybe you will like it.

Wouldn't it be easy for the game op to rig the table to have a extremely low winning ratio for a game like this? =/

Indeed. This means whoever creates it has a very good chance of making a significant amount of BTC. The players, not so much.

Not really. You'd just need to show an MD5 hash of the game outcome, before the bets are placed. Spoils the fun a bit of 'spinning' the wheel, but it's all digital anyway; no reason the random number can't be chosen in advance.

Wouldn't it be easy for the game op to rig the table to have a extremely low winning ratio for a game like this? =/

Indeed. This means whoever creates it has a very good chance of making a significant amount of BTC. The players, not so much.

Not really. You'd just need to show an MD5 hash of the game outcome, before the bets are placed. Spoils the fun a bit of 'spinning' the wheel, but it's all digital anyway; no reason the random number can't be chosen in advance.

hmm pardon my ignorance but how would this effect the control an op has over the win/lose ratio?

Wouldn't it be easy for the game op to rig the table to have a extremely low winning ratio for a game like this? =/

Indeed. This means whoever creates it has a very good chance of making a significant amount of BTC. The players, not so much.

Not really. You'd just need to show an MD5 hash of the game outcome, before the bets are placed. Spoils the fun a bit of 'spinning' the wheel, but it's all digital anyway; no reason the random number can't be chosen in advance.

hmm pardon my ignorance but how would this effect the control an op has over the win/lose ratio?

Because if the random number is chosen *before* bets are placed, and an MD5 hash of that number displayed, then the operator doesn't know what bets will be placed and whether or not the player is going to win or not.

1. System generates random number, 1-38 (or 37).2. Page displays some kind of hash, based on this number.3. Player places bets and submits4. System checks what they have won and shows the player exactly how the MD5 hash was generated so that the player can verify the number was chosen before bets were placed.

Wouldn't it be easy for the game op to rig the table to have a extremely low winning ratio for a game like this? =/

Indeed. This means whoever creates it has a very good chance of making a significant amount of BTC. The players, not so much.

Not really. You'd just need to show an MD5 hash of the game outcome, before the bets are placed. Spoils the fun a bit of 'spinning' the wheel, but it's all digital anyway; no reason the random number can't be chosen in advance.

hmm pardon my ignorance but how would this effect the control an op has over the win/lose ratio?

Because if the random number is chosen *before* bets are placed, and an MD5 hash of that number displayed, then the operator doesn't know what bets will be placed and whether or not the player is going to win or not.

1. System generates random number, 1-38 (or 37).2. Page displays some kind of hash, based on this number.3. Player places bets and submits4. System checks what they have won and shows the player exactly how the MD5 hash was generated so that the player can verify the number was chosen before bets were placed.

But if the op puts only 20 numbers in the random generation, he has decreased the win ratio and yet it still appears as if it's a random hash based on 38 numbers. I suppose there would need to be some way to display previous rolls...but even then...it seems so easy to put the odds in the op's favor.

Wouldn't it be easy for the game op to rig the table to have a extremely low winning ratio for a game like this? =/

Indeed. This means whoever creates it has a very good chance of making a significant amount of BTC. The players, not so much.

Not really. You'd just need to show an MD5 hash of the game outcome, before the bets are placed. Spoils the fun a bit of 'spinning' the wheel, but it's all digital anyway; no reason the random number can't be chosen in advance.

hmm pardon my ignorance but how would this effect the control an op has over the win/lose ratio?

Because if the random number is chosen *before* bets are placed, and an MD5 hash of that number displayed, then the operator doesn't know what bets will be placed and whether or not the player is going to win or not.

1. System generates random number, 1-38 (or 37).2. Page displays some kind of hash, based on this number.3. Player places bets and submits4. System checks what they have won and shows the player exactly how the MD5 hash was generated so that the player can verify the number was chosen before bets were placed.

But if the op puts only 20 numbers in the random generation, he has decreased the win ratio and yet it still appears as if it's a random hash based on 38 numbers. I suppose there would need to be some way to display previous rolls...but even then...it seems so easy to put the odds in the op's favor.

You could make it so that your data string contains all numbers, and the winning number is the first one... so you scramble the 38 numbers and hash the following string:

9,8,2,1,5,4,12,3,15... etc, etc etc (all 38 numbers)

The winning number is 9... after the game, you can show the player the md5 and also show them the data string which has 9 coming first.

You could still fake it so that 17 for example never came first, but over time I'm sure people would start to notice that 17 never wins.

Wouldn't it be easy for the game op to rig the table to have a extremely low winning ratio for a game like this? =/

Indeed. This means whoever creates it has a very good chance of making a significant amount of BTC. The players, not so much.

Not really. You'd just need to show an MD5 hash of the game outcome, before the bets are placed. Spoils the fun a bit of 'spinning' the wheel, but it's all digital anyway; no reason the random number can't be chosen in advance.

hmm pardon my ignorance but how would this effect the control an op has over the win/lose ratio?

Because if the random number is chosen *before* bets are placed, and an MD5 hash of that number displayed, then the operator doesn't know what bets will be placed and whether or not the player is going to win or not.

1. System generates random number, 1-38 (or 37).2. Page displays some kind of hash, based on this number.3. Player places bets and submits4. System checks what they have won and shows the player exactly how the MD5 hash was generated so that the player can verify the number was chosen before bets were placed.

But if the op puts only 20 numbers in the random generation, he has decreased the win ratio and yet it still appears as if it's a random hash based on 38 numbers. I suppose there would need to be some way to display previous rolls...but even then...it seems so easy to put the odds in the op's favor.

You could make it so that your data string contains all numbers, and the winning number is the first one... so you scramble the 38 numbers and hash the following string:

9,8,2,1,5,4,12,3,15... etc, etc etc (all 38 numbers)

The winning number is 9... after the game, you can show the player the md5 and also show them the data string which has 9 coming first.

You could still fake it so that 17 for example never came first, but over time I'm sure people would start to notice that 17 never wins.

I was thinking that too. The more you played it the more you'd be able to find out which 20 numbers or whatever were the ones in play. In that case, you could actually screw over the op pretty easily. Good times.

I was thinking that too. The more you played it the more you'd be able to find out which 20 numbers or whatever were the ones in play. In that case, you could actually screw over the op pretty easily. Good times.

The only way I can think of to prove all 38 are included would be to present the player with 38 buttons, each with a hash of a string which included the number, then have the player choose one of the 38 buttons, and that would be the winning number for the roulette bets.

Sounds complicated, but I'm not sure if there's simpler way for absolute proof.

I was thinking that too. The more you played it the more you'd be able to find out which 20 numbers or whatever were the ones in play. In that case, you could actually screw over the op pretty easily. Good times.

The only way I can think of to prove all 38 are included would be to present the player with 38 buttons, each with a hash of a string which included the number, then have the player choose one of the 38 buttons, and that would be the winning number for the roulette bets.

Sounds complicated, but I'm not sure if there's simpler way for absolute proof.

perhaps you could use the principle of inverse to solve this problem. 38 numbers, you choose 1. Instead of the computer actually playing the number(s) selected, it would play every number BUT that one. That way, if it actually lands on the spot where nothing is placed, it knows to pay out the user.

I was thinking that too. The more you played it the more you'd be able to find out which 20 numbers or whatever were the ones in play. In that case, you could actually screw over the op pretty easily. Good times.

The only way I can think of to prove all 38 are included would be to present the player with 38 buttons, each with a hash of a string which included the number, then have the player choose one of the 38 buttons, and that would be the winning number for the roulette bets.

Sounds complicated, but I'm not sure if there's simpler way for absolute proof.

perhaps you could use the principle of inverse to solve this problem. 38 numbers, you choose 1. Instead of the computer actually playing the number(s) selected, it would play every number BUT that one. That way, if it actually lands on the spot where nothing is placed, it knows to pay out the user.

Wait...I'm not sure that made sense at all. hah

I didn't quite understand, no.

Also, when I said 38 buttons, I meant in addition to the roulette betting table, not as a replacement for it.